求助利用matlab计算含三角函数的方程组的数值解,如何使数值解更为精确?13576000*(cos(x(1))+cos(a)-(1+b/2)^2*(1-b)*(cos(a)+cos(x(2))))-x(3)*sin(c)=013576000*(sin(x(1))+(1+b/2)^2*(1-b)*sin(x(2)))-x(3)*cos(c)=0x(1)-(1+b/2)*(1-b)*
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![求助利用matlab计算含三角函数的方程组的数值解,如何使数值解更为精确?13576000*(cos(x(1))+cos(a)-(1+b/2)^2*(1-b)*(cos(a)+cos(x(2))))-x(3)*sin(c)=013576000*(sin(x(1))+(1+b/2)^2*(1-b)*sin(x(2)))-x(3)*cos(c)=0x(1)-(1+b/2)*(1-b)*](/uploads/image/z/5434680-48-0.jpg?t=%E6%B1%82%E5%8A%A9%E5%88%A9%E7%94%A8matlab%E8%AE%A1%E7%AE%97%E5%90%AB%E4%B8%89%E8%A7%92%E5%87%BD%E6%95%B0%E7%9A%84%E6%96%B9%E7%A8%8B%E7%BB%84%E7%9A%84%E6%95%B0%E5%80%BC%E8%A7%A3%2C%E5%A6%82%E4%BD%95%E4%BD%BF%E6%95%B0%E5%80%BC%E8%A7%A3%E6%9B%B4%E4%B8%BA%E7%B2%BE%E7%A1%AE%3F13576000%2A%28cos%28x%281%29%29%2Bcos%28a%29-%281%2Bb%2F2%29%5E2%2A%281-b%29%2A%28cos%28a%29%2Bcos%28x%282%29%29%29%29-x%283%29%2Asin%28c%29%3D013576000%2A%28sin%28x%281%29%29%2B%281%2Bb%2F2%29%5E2%2A%281-b%29%2Asin%28x%282%29%29%29-x%283%29%2Acos%28c%29%3D0x%281%29-%281%2Bb%2F2%29%2A%281-b%29%2A)
求助利用matlab计算含三角函数的方程组的数值解,如何使数值解更为精确?13576000*(cos(x(1))+cos(a)-(1+b/2)^2*(1-b)*(cos(a)+cos(x(2))))-x(3)*sin(c)=013576000*(sin(x(1))+(1+b/2)^2*(1-b)*sin(x(2)))-x(3)*cos(c)=0x(1)-(1+b/2)*(1-b)*
求助利用matlab计算含三角函数的方程组的数值解,如何使数值解更为精确?
13576000*(cos(x(1))+cos(a)-(1+b/2)^2*(1-b)*(cos(a)+cos(x(2))))-x(3)*sin(c)=0
13576000*(sin(x(1))+(1+b/2)^2*(1-b)*sin(x(2)))-x(3)*cos(c)=0
x(1)-(1+b/2)*(1-b)*(pi-x(2))=0
利用[x,fv.ef.out.jac]=fsolve(@NSC,x0,[],pi/12,0.1,88*pi/180)
初值设为
0 ,3 ,70000000
得到的数值解不够精确,理论上x(1)应该略大于0,x(2)接近3.14,可惜得不到,
我是matlab菜鸟,如何进行次方程组的数值计算!
M文件:
function y=NSC(x,a,b,c)
y(1)=13576000*(cos(x(1))+cos(a)-(1+b/2)^2*(1-b)*(cos(a)+cos(x(2))))-x(3)*sin(c);
y(2)=13576000*(sin(x(1))+(1+b/2)^2*(1-b)*sin(x(2)))-x(3)*cos(c);
y(3)=x(1)-(1+b/2)*(1-b)*(pi-x(2));
matlab命令:
x0=[0,3,7e7]
[x,fv.ef.out.jac]=fsolve(@NSC,x0,[],pi/12,0.1,88*pi/180)
求助利用matlab计算含三角函数的方程组的数值解,如何使数值解更为精确?13576000*(cos(x(1))+cos(a)-(1+b/2)^2*(1-b)*(cos(a)+cos(x(2))))-x(3)*sin(c)=013576000*(sin(x(1))+(1+b/2)^2*(1-b)*sin(x(2)))-x(3)*cos(c)=0x(1)-(1+b/2)*(1-b)*
这个数字精度应该是可以设置的,float型或者是double型都可以,这样就不会取整了
cuoleba